Elliptic functions lecture 4. The sigma function

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  • Опубликовано: 30 янв 2025
  • This lecture is part of a series of lectures on elliptic functions.
    It covers the sigma function, the simplest theta function, and discusses line bundles over C/L.
    For the other lectures in the course see
    • Elliptic functions

Комментарии • 15

  • @debblez
    @debblez 11 месяцев назад +13

    this series is what I was missing in my life

  • @mrlimemil
    @mrlimemil 11 месяцев назад +1

    Such a great lecture series.

  • @JAYMOAP
    @JAYMOAP 11 месяцев назад +5

    Thank you Richard much appreciated

  • @Juniper-111
    @Juniper-111 11 месяцев назад +1

    at 0:30 you say that the Weierstrauss p function has 2 poles on the fundamental domain and the Jacobi functions have 1 pole on the fundamental domain which seems to be backwards
    edit: I realized my mistake. In the last video, we modified the fundamental domain but changed the our condition on the periodicity so that f(z+lambda)=\pm f(z).
    edit 2: I imagine that by saying the Weierstrauss p function has 2 poles, we really mean that it has one double pole and that this forces the function to have two zeroes in the fundamental domain by the argument principle

  • @JAYMOAP
    @JAYMOAP 11 месяцев назад

    Keep up the good work Richard. Take a look at mixed boundary value problem with electrostatic disk with hole/s. I try to contribute as much as I can we can narrow this down my friend :)

  • @KoHaN7
    @KoHaN7 11 месяцев назад

    Extremely interesting content as always! Thank you for makin these series.
    My first thoughs when trying to create a periodic function thetha(z) with zeros at the lattice, with the Weierstrass P function with poles of order 2, would have been to start with thetha(z) ~1/sqrt(P(z)). In the video is follows that P(z) = -log(thehta(z)) ' ', whereas in my case it would be P(z) ~ 1/thetha(z)^2. I understand that the second identiy is wrong because it makes P(z) no longer periodic since thehta(z) is not exactly periodic, but I feel maybe trying to correct that discrepancy, instead of doing the integral twice, could be an alternative deffinition?
    Could someone tell me where my assumptinos went wrong?
    Edit:
    After watching the rest of the video, I now understand that the P(z) should in fact be expressable by a ratio of products of thetha. Somthing like P~ thehta(z-zero1) thehta(z-zero2) /theha(z)^2 in the fundamental domain. By Liouvile's theorem, there should be a constant of proportionality between these two functions. Therefore you have an implicit definition of thetha(z) as function of P(z)? I wonder if it is easy to invert...
    Edit2:
    I continued down this rabit hole, and started considering combining what we know. For simplicity I call thetha = T
    Since P(z) = - log(T(z)) ' ' , Therefore using the second definiton, withouth knowing the constant
    log(P(z)) = log(T(z-zero1)) + log(T(z-zero2)) - 2 log(T(z)) + Constant, Differentiating twice
    log(P(z)) ' ' = log(T(z-zero1))' ' + log(T(z-zero2)) ' ' - 2 log(T(z)) ' ', Using the first identity
    log(P(z)) ' ' = - P(z-zero1) - P(z-zero2) + 2 P(z)
    This is an interesting functional equation however I have not been able to find anything relatd to it, has anyone seen a similar identity before? Is it really an Identity? I am aware that importnt constants have been ommited and I might have been sloppy with my algebra, so I might be wrong of course, but it is still fun to think about it!

  • @hypercube_0
    @hypercube_0 7 месяцев назад

    Hello, Can i use sigma function to prove the infinite product of lemniscate sin ? Thank you !

  • @vadymfedyukovych8660
    @vadymfedyukovych8660 11 месяцев назад +1

    This covers lots of gaps from "easy pre-requisite" and introductions into elliptic curves cryptography.

  • @SeanEberhard
    @SeanEberhard 11 месяцев назад

    If you have a 1-cocyle c_\lambda(z), is it guaranteed that there is a function f(z) satisfying f(z+\lambda) = c_\lambda(z) f(z)? (It is given that c_{\lambda+\mu}(z) = c_\lambda(z) c_\mu(z+\lambda), so it is tempting to sum over \mu, but that seems very unlikely to converge.)

  • @JAYMOAP
    @JAYMOAP 11 месяцев назад +2

    Yes it's theta

  • @JAYMOAP
    @JAYMOAP 11 месяцев назад +1

    Laplace beltrami

  • @JohnSmith-ux3xm
    @JohnSmith-ux3xm 11 месяцев назад +1

    Oh dear. It has gone very tricky to understand.

  • @kingarth0r
    @kingarth0r 11 месяцев назад +3

    ohio sigma rizz

  • @aidanyoung9133
    @aidanyoung9133 11 месяцев назад

    S I G M A

  • @alexlouismusic
    @alexlouismusic 11 месяцев назад

    The LIGMA function